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Consider a positive stochastic process $X_t$ defined on probability space $(\Omega,\mathcal{F},\mathbb{P})$. The process converges almost surely to a deterministic limit $X$ (may be infinite). One connection between almost sure convergence of $X_t$ and convergence in mean is given by Fatou's lemma. Given that the limit of the mean exists, the lemma implies

$$\mathbb{E}[\liminf_{t \to \infty}X_t] \leq \lim_{t \to \infty} \mathbb{E}[X_t]$$

Moreover, $\liminf_{t \to \infty}X_t(\omega)$ is $X$ for all $\omega\in \Omega$ except for those in a set of measure zero. For those $\omega$ in a set of measure zero, $\liminf_{t \to \infty}X_t(\omega)$ is still bounded below by zero. Hence

$$X \leq \mathbb{E}[\liminf_{t \to \infty}X_t] \leq \lim_{t \to \infty} \mathbb{E}[X_t]$$

Therefore the almost sure limit bounds the limit of the mean from below.

Is the argument correct? Are there other connections between almost sure limits and limits of the mean? (particularly interested in positive processes)

fes
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  • Probably the most important connection in general is that a sequence convergent in measure has a subsequence convergent almost everywhere. I'm not quite sure what this implies for stochastic processes. – tomasz Jun 03 '17 at 09:21
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    The statement after "due to the positivity of $X_t$" doesn't make any sense to me. Just because $\mathbb{E}[X]$ is less than some number, doesn't mean you can make the same conclusion about $X$ itself. $X$ could be much larger than its expectation in some outcomes (though they would necessarily have small probability). – Nate Eldredge Jun 03 '17 at 14:41
  • @Nate Eldredge. I changed the question a bit. The idea is that if the limit exists it is X in a set of measure one and a number that is bounded from below in set of measure zero. Do you think there is a way to make this more formal by introducing an additional assumption etc.? – fes Jun 03 '17 at 15:02
  • @tomasz Thank you. Does my result make sense to you? See the post by Nate Eldredge. – fes Jun 03 '17 at 15:09
  • Are you assuming that $X$ is deterministic? – Did Jun 03 '17 at 15:24
  • @Did. I have a stochastic process $X_t$ that converges almost surely to a deterministic number $X$ or infinity. The process is positive. Do I need an additional assumption for the inequality to work? Thank you. – fes Jun 03 '17 at 15:30
  • So, you are assuming that $X$ is deterministic. – Did Jun 03 '17 at 15:31
  • @ Yes. The limit $X$ is deterministic. – fes Jun 03 '17 at 15:35
  • @Did I changed the limit of $X_t$ to liminf. In my case the limit of the mean exists always. – fes Jun 03 '17 at 15:45
  • @Did. Dear did, I clarified the argument. It seems correct to me. Is there still some hole I am unable to see? – fes Jun 04 '17 at 08:25
  • It is correct, and one can simplify it somewhat, noting that, since one assumes that $(X_t)$ converges almost surely, one has $$E(\liminf X_t)=E(\lim X_t)=X$$ Recall that if $Y=Z$ almost surely and if $Y$ is integrable then $Z$ is integrable and $E(Z)=E(Y)$. – Did Jun 04 '17 at 09:44
  • Another well-known "connection between almost sure convergence and convergence in mean" is that almost sure convergence implies convergence in probability and that convergence in probability + uniform integrability imply $L^1$ convergence, hence almost sure convergence + uniform integrability imply $L^1$ convergence. – Did Jun 04 '17 at 09:47
  • @Did Thank you. So do we need the positivity of $X_t$ at all? – fes Jun 04 '17 at 10:06
  • For Fatou, yes. For the implication in my previous comment, no. Say, which text are you using on these matters? – Did Jun 04 '17 at 10:07
  • @Did. Ok got it. I have e.g. Durrett's book. – fes Jun 04 '17 at 10:12
  • @Did. I have seen the first result somewhere (almost surely equal processes with one integrable implies mean/L1 equivalence). Could you give me any book to cite? – fes Jun 04 '17 at 10:26

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